Two cells of internal resistance `r_1 and r_2` and of same emf are connected in series across a resistor of resistance R. If the terminal potential difference across the call of internal resistance `r_1` is zero , then the value of R is

Options

A. `R=2(r_(1)+r_(2))`
B. `R=r_(2)-r_(1)`
C. `R=r_(1)-r_(2)`
D. `R=2(r_(1)-r_(2))`

Answer

C

Explanation

Current in the circuit : `I=(2E)/(R+r_(1)+r_(2))`
Terminal p.d across 1 st cell is `V_(1)=E-Ir_(1)`
Given : `V_(1)=0`
`rArrE-Ir_(1)=0`
`E-((2E)/(R+r_(1)+r_(2)))r_(1)=0`
`E=(2Er_(1))/(R+r_(1)+r_(2))`
`R+r_(1)+r_(2)=2r_(1)`
or `R=r_(1)-r_(2)`